Angle between tangent drawn from the point P(1, – 1) to the circle x2 + y2 + 8x + 6y = 0 is
Answers
Given : point P(1, – 1) and circle x2 + y2 + 8x + 6y = 0 is
To find : Angle between tangent drawn from the point P(1, – 1)
Solution:
x² + y² + 8x + 6y = 0
=> (x + 4)² - 16 + (y + 3)² - 9 = 0
=> (x + 4)² + (y + 3)² = 25
=> (x + 4)² + (y + 3)² = 5²
=> Center = - 4 , - 3
Radius = 5
Distance Between P (1 , - 1) & center (-4 ,3 )
= √(-4 - 1)² + (3 -(-1))²
= √25 + 16
= √41
2x is the angle between tangent drawn from the point P(1, – 1) to the circle x²² + y² + 8x + 6y = 0 is
Sin x = 5/√41
x =Sin⁻¹(5/√41)
=> x = 51.34°
=> 2x = 102.68°
Angle between tangent drawn from the point P(1, – 1) to the circle x² + y² + 8x + 6y = 0 is 102.68°
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